Control Systems I. Lecture 4: Diagonalization, Modal Analysis, Intro to Feedback. Readings: Emilio Frazzoli


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1 Control Systems I Lecture 4: Diagonalization, Modal Analysis, Intro to Feedback Readings: Emilio Frazzoli Institute for Dynamic Systems and Control DMAVT ETH Zürich October 13, 2017 E. Frazzoli (ETH) Lecture 4: Control Systems I 13/10/ / 26
2 Tentative schedule # Date Topic 1 Sept. 22 Introduction, Signals and Systems 2 Sept. 29 Modeling, Linearization 3 Oct. 6 Analysis 1: Time response, Stability 4 Oct. 13 Analysis 2: Diagonalization, Modal coordinates. 5 Oct. 20 Transfer functions 1: Definition and properties 6 Oct. 27 Transfer functions 2: Poles and Zeros 7 Nov. 3 Analysis of feedback systems: internal stability, root locus 8 Nov. 10 Frequency response 9 Nov. 17 Analysis of feedback systems 2: the Nyquist condition 10 Nov. 24 Specifications for feedback systems 11 Dec. 1 Loop Shaping 12 Dec. 8 PID control 13 Dec. 15 Implementation issues 14 Dec. 22 Robustness E. Frazzoli (ETH) Lecture 4: Control Systems I 13/10/ / 26
3 Recap of the previous lecture LTI system: ẋ(t) = Ax(t) + Bu(t), y(t) = Cx(t) + Du(t). Time response: x(t) = e At x 0 + y(t) = Ce At x 0 + C t 0 t 0 e A(t τ) Bu(τ) dτ, e A(t τ) Bu(τ) dτ + Du(t). Easy to compute if the matrix A is diagonal, in which case: x i (t) = e λ i t x 0,i + y(t) = t 0 e λ i (t τ) b i u(τ) dτ, n c i x i (t) + Du(t). i=1 The eigenvalues of A, λ i, can be real or complexconjugate, giving rise to simple exponentials, or oscillations with exponentially changing magnitude, respectively. Asymptotically stable if Re(λ i ) < 0 for all i. E. Frazzoli (ETH) Lecture 4: Control Systems I 13/10/ / 26
4 Today s learning objectives After today s lecture, you should be able to: Diagonalize a matrix using similarity transformations. Describe the behavior of an LTI system in modal coordinates. Understand concepts like controllability and observability. Understand the basic effects of feedback control on the closedloop dynamics. E. Frazzoli (ETH) Lecture 4: Control Systems I 13/10/ / 26
5 Similarity Transformations The choice of a statespace model for a given system is not unique. For example, let T be an invertible matrix, and consider a coordinate transpormation x = T x, i.e., x = T 1 x. This is called a similarity transformation. The standard statespace model can be written as { ẋ = Ax + Bu, y = Cx + Du. { T x = AT x + Bu, y = CT x + Du. i.e., x = (T 1 AT ) x + (T 1 B)u = Ã x + Bu y = (CT ) x + Du = C x + Du. You can check that the time response is exactly the same for the two models (A, B, C, D) and (Ã, B, C, D)! E. Frazzoli (ETH) Lecture 4: Control Systems I 13/10/ / 26
6 Diagonalization Let λ i, v i be respectively an eigenvalue and an eigenvector of A, i.e., Av i = λ i v i. v n λ n v n Now assume we have n (=dim. of x and A) independent eigenvectors; then we can assemble the eigenvectors into an invertible matrix V whose columns are the eigenvectors v i. Then AV = A v 1 v 2... = λ 1 v 1 λ 2 v 2... = V Λ. In other words, if a square matrix A has a full set of independent eigenvectors, then it is diagonalizable (and viceversa), with the similarity transformation given by a matrix whose columns are the eigenvectors. E. Frazzoli (ETH) Lecture 4: Control Systems I 13/10/ / 26
7 Modal decomposition The entries in the diagonal matrix Ã = Λ are the eigenvalues λ 1,..., λ n of the matrix A. Since x(t) = eãt x(0), we get that each component of the homogeneous solution is given by x i (t) = e λ i t x i (0). Furthermore, if x(0) = v i for some i = 1,..., n, then by the definition of matrix exponential and eigenvalues/eigenvectors x(t) = e At v i = e λ i t v i. If the eigenvectors v i, i = 1,..., n form a basis, then we can always express any initial condition as a linear combination of eigenvectors, i.e., x(0) = V x(0), with x 0 = V 1 x 0, and write x(t) = n e λ i t x i (0) v i. i=1 E. Frazzoli (ETH) Lecture 4: Control Systems I 13/10/ / 26
8 Modal coordinates Eigenvalues and eigenvectors of A define the modes of the system; the transformed coordinates x = Vx are also called the modal coordinates. The eigenvector v i defines the shape of the ith mode; The modal coordinate x i scales the mode (e.g., at the initial condition); The eigenvalue λ i defines how the amplitude of the mode evolves over time. 1 As an exponential e λ i t x 0 for real λ i 2 As an sinusoid with exponentially changing amplitude e σt sin(ωit + φ 0 )x 0 for complexconjugate λ i = σ i + jω i. 3 As polynomiallyscaled versions of the above t p e λ i t for repeated λ i. The amplitude of the i mode goes to zero as t increases if and only if Re(λ i ) < 0. E. Frazzoli (ETH) Lecture 4: Control Systems I 13/10/ / 26
9 Example: Simple Pendulum Recall the model for a simple pendulum, assuming no damping (and no input/output for now): [ ] 0 1 ẋ = x g/l 0 Eigenvalues are solutions of the characteristic equation: [ ] λ 1 det(λi A) = det = λ 2 + g/l = 0, g/l λ i.e., g λ 1,2 = ±j l. Eigenvectors are obtained as solutions of (λi A)v = 0, e.g., [ ] 1 v 1,2 = ±j g l E. Frazzoli (ETH) Lecture 4: Control Systems I 13/10/ / 26
10 Pendulum mode shape Assume x(0) = (1, 0), i.e., x(0) = (1/2, 1/2). Im Im Re after time t: multiply by e λ i t Re Combining the two modes, we get x 1 (t) = cos ωt, x 2 (t) = ω sin ωt. E. Frazzoli (ETH) Lecture 4: Control Systems I 13/10/ / 26
11 Cartesian vs. polar form of complex numbers Complex numbers can be represented in basically two ways: Cartesian form: z = a + jb Polar form: z = z e j z In the above formulas, z = a 2 + b 2, and z = arctan(b/a) (arctan understood as the fourquadrant version). The Cartesian form makes addition easy: (a 1 + jb 1 ) + (a 2 + jb 2 ) = a 1 + a 2 + j(b 1 + b 2 ) The polar form makes multiplication easy: m 1 e jφ1 m 2 e jφ2 = (m 1 + m 2 )e j(φ1+φ2) E. Frazzoli (ETH) Lecture 4: Control Systems I 13/10/ / 26
12 General evolution for complexconjugate modes Im Im Re after time t: multiply by e λ i t Re Combining the two modes, we get x 1 (t) = c 1 e σt sin(ωt + φ 1, x 2 (t) = c 2 e σt sin(ωt + φ 2 ). E. Frazzoli (ETH) Lecture 4: Control Systems I 13/10/ / 26
13 Towards feedback control So far we have looked at how a given system, represented as a statespace model or as a transfer function, behaves given a certain input (and/or initial condition). Typically the system behavior may not be satisfactory (e.g., because it is unstable, or too slow, or too fast, or it oscillates too much, etc.), and one may want to change it. This can only be done by feedback control! However, we need to understand to what extent we can change the system behavior. More precisely, for control design, one must also understand how the control input can affect the state of the system; how the state of the system affects the output. E. Frazzoli (ETH) Lecture 4: Control Systems I 13/10/ / 26
14 Controllability and Observability An LTI system of the form ẋ = Ax + Bu is said to be controllable if for any given initial state x(0) = x c there exists a control signal that takes the state to the origin x(t) = 0 for some finite time t. An LTI system of the form ẋ = Ax + Bu, y = Cx + Du is said to be observable if any given initial condition x(0) = x o can be reconstructed based on the knowledge of the input and output signal only, over a finite time interval [0, t]. E. Frazzoli (ETH) Lecture 4: Control Systems I 13/10/ / 26
15 Intuition from the Modal Form Recall that if the matrix A has a complete set of independent (right) eigenvectors {v 1,..., v n }, with eigenvalues {λ 1,..., λ n } it can be diagonalized by the matrix T = [ v 1 v 2... v n ]. The transformed state in the diagonalized system is such that x = T x. The transformed model is (Ã, B, C, D) = (T 1 AT, T 1 B, CT, D). Componentwise, the dynamics of each modal coordinate x i are given by The output is given by d dt x i(t) = λ i x i (t) + b i u(t), i = 1,..., n. y = c 1 x 1 (t) + c 2 x 2 (t) c n x n (t) + Du(t). E. Frazzoli (ETH) Lecture 4: Control Systems I 13/10/ / 26
16 Controllability/observability for diagonal systems From the previous slide, we can deduce The ith coordinate can be controlled if and only if b i 0. The ith coordinate appears in the output if and only if c i 0. Hence: An LTI system in diagonal form is controllable if b i 0, i = 1,..., n. An LTI system in diagonal form is observable if c i 0, i = 1,..., n. An LTI system is stabilizable if all unstable modes are controllable. An LTI system is detectable if all unstable modes are observable. E. Frazzoli (ETH) Lecture 4: Control Systems I 13/10/ / 26
17 An example 1.618j Ã = 1.618j 0.618j j, B = j j C = [ 0.276j 0.276j ], D = [0]. E. Frazzoli (ETH) Lecture 4: Control Systems I 13/10/ / 26
18 An example j Ã = j j 0.707j, B = 0.707j 0 j 0 C = [ ], D = [0]. E. Frazzoli (ETH) Lecture 4: Control Systems I 13/10/ / 26
19 More general conditions Consider the derivatives of the output: y(0) = Cx(0), y (0) = CAx(0), y (0) = CA 2 x(0),... One can reconstruct x(0), i.e., the system is observable, as long as the observability matrix C CA CA 2... CA n 1 has full rank. Note that it is sufficient to compute the observability matrix only up to the power n 1 of A (CayleyHamilton theorem). A similar condition can be obtained for controllability, with a somewhat more complicated proof: a system is controllable if the controllability matrix [ B AB A 2 B... A n 1 B ] has full rank. E. Frazzoli (ETH) Lecture 4: Control Systems I 13/10/ / 26
20 Pole placement first order systems Consider a firstorder control system with dynamics ẋ = ax + bu, and assume that we are not happy about its behavior (e.g., it is unstable since a > 0, or maybe stable but slow because a is small). Can we change the behavior by choosing u in a clever way? Feedback control: choose u = kx; then the dynamics would become ẋ = (a bk)x As long as b 0 (i.e., if the system is controllable), by choosing k = (a a )/b, we can place the closedloop eigenvalue at a desired value a, or anywhere we want on the real axis! This is the simplest example of a general technique called pole placement. E. Frazzoli (ETH) Lecture 4: Control Systems I 13/10/ / 26
21 Example Assume ẋ = 0.5x + u, i.e., the system is unstable with time constant 2. We would like the system to be stable, with time constant 1/2. Choose u = ( )x = 2.5x; the closed loop will be ẋ = 2x as desired. openloop closedloop u t E. Frazzoli (ETH) Lecture 4: Control Systems I 13/10/ / 26
22 Effect on feedback for closedloop dynamics If we have an openloop LTI system ẋ(t) = Ax(t) + Bu(t), y(t) = Cx(t), by choosing a linear feedback u = ky = kcx, we can transform it into another, closedlopp LTI system: ẋ(t) = (A BkC)x(t), y(t) = Cx(t), In general negative feedback (i.e., u = ky ) has stabilizing effects, and the bigger k, the faster the closedloop system is, and the smaller the errors are. However this is not generally the case. In the rest of the course, we will look at ways to analyze the behavior of the closedloop system, and choosing the feedback control law, without necessarily lots of computation but rather using primarily graphical methods. E. Frazzoli (ETH) Lecture 4: Control Systems I 13/10/ / 26
23 Examples ẋ(t) = 0.5x(t) + u(t), y(t) = x(t), 0.3 Root Locus 0.2 Imaginary Axis (seconds 1 ) Real Axis (seconds 1 ) E. Frazzoli (ETH) Lecture 4: Control Systems I 13/10/ / 26
24 Examples ẋ(t) = [ ] [ ] x(t) + u(t), y(t) = [ ] x(t), 2.5 Root Locus Imaginary Axis (seconds 1 ) Real Axis (seconds 1 ) E. Frazzoli (ETH) Lecture 4: Control Systems I 13/10/ / 26
25 Examples ẋ(t) = x(t) + 0 u(t), y(t) = [ ] x(t), 4 Root Locus 3 2 Imaginary Axis (seconds 1 ) Real Axis (seconds 1 ) E. Frazzoli (ETH) Lecture 4: Control Systems I 13/10/ / 26
26 Today s learning objectives After today s lecture, you should be able to: Diagonalize a matrix using similarity transformations. Describe the behavior of an LTI system in modal coordinates. Understand concepts like controllability and observability. Understand the basic effects of feedback control on the closedloop dynamics. E. Frazzoli (ETH) Lecture 4: Control Systems I 13/10/ / 26
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